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![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x. (x2 – 7x -3)
= 2x3- 14x2- 6x
b) ( -2x3 + y2 -7xy). 4xy2
= -8x4y2+ 4xy4- 28x2y3
c)(-5x3).(2x2+3x-5)
= -10x5-15x4+25x3
d) (2x2 - xy+ y2).(-3x3)
=-6x5+ 3x4y -3x3y2
e)(x2 -2x+3). (x-4)
=x3-2x2+3x -4x2+8x-12
=x3-6x2+11x-12
f) ( 2x3 -3x -1). (5x+2)
=10x4-15x2-5x +4x3-6x-2
=10x4+4x3-15x2-11x-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2x\left(x^2-7x-3\right)=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)\cdot4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-4x^2-2x^2+8x+3x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\cdot\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)